1.

A ray of light refracts from medium 1 into a thin layer of medium 2, crosses the layer and is incident at the critical angle on the interface between the medium 2 and 3 as shown in the figure. If the angle of incidence of ray is theta, the values of theta is

Answer»

`sin^(-1)((8)/(9))`
`sin^(-1)((13)/(18))`
`sin^(-1)((13)/(16))`
`sin^(-1)((8)/(13))`

Solution :Light enters from medium 1 to medium 2
`(sin theta)/(sin r) = ""_(1)mu_(2)=(mu_(2))/(mu_(1))`...(A)
then again light falls from medium 2 to medium 3 at critical angle
`( sin r)/(sin 90^(@))= ""_(2)mu_(3)= (mu_(3))/(mu_(2))`
`sin r = (1.3)/(1.8)`
Put the value in equation (A) we get,
`sin theta = (1.8)/(1.6)xx(1.3)/(1.8)`
`rArr theta= sin^(-1) (13/16)`


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